Problem: Is ${605097}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {605097}= &&{6}\cdot100000+ \\&&{0}\cdot10000+ \\&&{5}\cdot1000+ \\&&{0}\cdot100+ \\&&{9}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {605097}= &&{6}(99999+1)+ \\&&{0}(9999+1)+ \\&&{5}(999+1)+ \\&&{0}(99+1)+ \\&&{9}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {605097}= &&\gray{6\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {6}+{0}+{5}+{0}+{9}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${605097}$ is divisible by $3$ if ${ 6}+{0}+{5}+{0}+{9}+{7}$ is divisible by $3$ Add the digits of ${605097}$ $ {6}+{0}+{5}+{0}+{9}+{7} = {27} $ If ${27}$ is divisible by $3$ , then ${605097}$ must also be divisible by $3$ ${27}$ is divisible by $3$, therefore ${605097}$ must also be divisible by $3$.